Question

Suppose a 32pound weight stretches a spring 2feet. If the weight is released from rest at the equilibrium position, find the equation of motion$x\left(t\right)$if an impressed force$f\left(t\right)=20t$acts on the system for$0⩽t<5$and is then removed (see Example 5). Ignore any damping forces. Use a graphing utility to graphdata-custom-editor="chemistry" $x\left(t\right)$on the interval$[0,10]$.

Short answer

The equation of motion $x\left(t\right)$if an impressed force $f\left(t\right)=20t$ acts on the system for $0⩽t<5$and is then removed is

.

$$\begin{gathered} x\left(t\right)=\frac{5}{4}t-\frac{5}{16}\sin \left(4t\right)+\frac{25}{4}U(t-5)-\frac{5}{4}(t-5)U(t-5) \\ -\frac{25}{4}\cos \left(4\right(t-5\left)\right)U(t-5)+\frac{5}{16}\sin \left(4\right(t-5\left)\right)U(t-5) \end{gathered}$$

The graph of $x\left(t\right)$is

Step-by-step solution

Define Laplace transform.

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. The Laplace transform $\mathcal{L}$of a function $f\left(t\right)$ is defined as

$\mathcal{L}\left\{f\right(t\left)\right\}=\int {}_0{}^{\mathrm{\mathbb{N}}}{e}^{-st}f\left(t\right)dt$

In words, we can describe this expression as the Laplace transform of $f\left(t\right)$equals function F of S.

$\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$

Find the weight that's released from rest.

Weight,$w=32lb$

$32lb$weight stretches the spring 2 feet.

Using Hooke’s law, F = ks

$$\begin{gathered} 32=k·2 \\ k=16\text{lb}/\text{ft} \end{gathered}$$

We know that$m=\frac{w}{g}$, where$g=32\text{ft}/{\text{sec}}^2$ is the acceleration due to gravity.

Substituting$w=32lb$and$g=32\text{ft}/{\text{sec}}^2$ , we get

$m=\frac{w}{g}=\frac{32}{32}=1$

The weight is released from equilibrium point it represents $x\left(0\right)=0$. This is an initial condition.

Also the weight is released from rest, it represents$x\left(0\right)=0$

Find the Laplace transform for the given expression.

The force function is

$$\begin{gathered} f\left(t\right)=\left\{\begin{array}{ll}20t& 0⩽t<5\\ 0& t⩾5\end{array}\right. \\ =20t-20tU(t-5) \end{gathered}$$

The general equation for a spring mass system with an external force is:

$mx\text{'}\text{'}+kx=f\left(t\right)$

Substitute $m=1,k=16\text{and}f\left(t\right)=20t-20tU(t-5)$in the above equation.

$x\text{'}\text{'}+16x=20t-20tU(t-5)$

Take the Laplace transform on both sides,

$$\begin{gathered} \mathcal{L}\{20t-20tU(t-5\left)\right\}=\mathcal{L}\left\{x\text{'}\text{'}+16x\right\} \\ =s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x\text{'}\left(0\right)+16\mathcal{L}\left\{x\right\} \\ =\left(s^2+16\right)\mathcal{L}\left\{x\right\}\left(x\left(0\right)=0,x\text{'}\left(0\right)=0\right) \\ \mathcal{L}\left\{x\text{'}\text{'}\right\}=s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x\text{'}\left(0\right) \end{gathered}$$

Replace $\mathcal{L}\left\{x\right\}$by $X\left(s\right)$

$\mathcal{L}\{20t-20tU(t-5\left)\right\}=\left(s^2+16\right)X\left(s\right)$

The Laplace transform for$$\begin{gathered} x\text{'}\text{'}+16x=20t-20tU(t-5)is \\ \mathcal{L}\{20t-20tU(t-5\left)\right\}=\left(s^2+16\right)X\left(s\right) \end{gathered}$$

Find the value

Using the below equation we find the value of $\mathcal{L}\{20t-20tU(t-5\left)\right\}$.

If and then

$F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}anda>0then,\mathcal{L}\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

$\mathcal{L}\left\{U\right(t-a\left)\right\}=\frac{e^{-as}}{s}$

localid="1663942604189" $$\begin{gathered} \mathcal{L}\{20t-20tU(t-5\left)\right\}=20\mathcal{L}\left\{t\right\}-20\mathcal{L}\left\{tU\right(t-5\left)\right\} \\ =20\frac{1}{s^2}-20\mathcal{L}\left\{\right(t-5+5\left)U\right(t-5\left)\right\} \\ =20\frac{1}{s^2}-20\mathcal{L}\left\{\right(t-5\left)U\right(t-5\left)\right\}+100\mathcal{L}\left\{\right(U(t-5)\} \\ =20\frac{1}{s^2}-20\frac{1}{s^2}{e}^{-5s}+\frac{100}{s}{e}^{-5s} \end{gathered}$$

Hence the value of$\mathcal{L}\{20t-20tU(t-5\left)\right\}$is$20\frac{1}{s^2}-20\frac{1}{s^2}{e}^{-5s}+\frac{100}{s}{e}^{-5s}$

Substituting the Equation

Substitution of equations:

$$\begin{gathered} \left(s^2+16\right)X\left(s\right)=\frac{20}{s^2}-\frac{20}{s^2}{e}^{-5s}+\frac{100}{s}{e}^{-5s} \\ X\left(s\right)=\frac{20}{s^2\left(s^2+16\right)}+\frac{e^{-5s}(100s-20)}{s^2\left(s^2+16\right)} \end{gathered}$$

Decompose $\frac{20}{s^2\left(s^2+16\right)}$using partial fraction technique.

$$\begin{gathered} \frac{20}{s^2\left(s^2+16\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+16} \\ =\frac{As\left(s^2+16\right)+B\left(s^2+16\right)+(Cs+D)s^2}{s^2\left(s^2+16\right)} \end{gathered}$$

Since the fractions have the same denominators, it takes their numerators to be equal.

$$\begin{gathered} 20=A{s}^3+16As+B{s}^2+16B+C{s}^3+D{s}^2 \\ 20=(A+C)s^3+(B+D)s^2+16As+16B \end{gathered}$$

The decomposed value of the given equation is

$20=(A+C)s^3+(B+D)s^2+16As+16B$

Comparing the coefficients

Compare the coefficients on both sides of $20=(A+C)s^3+(B+D)s^2+16As+16B$

$$\begin{gathered} A+C=0 \\ B+D=0 \\ 16A=0 \\ 16B=20 \\ C=-A \\ D=-B \\ A=0 \\ B=\frac{5}{4} \\ C=0 \\ D=-\frac{5}{4} \\ A=0 \\ B=\frac{5}{4} \end{gathered}$$

Substitute the value of$A,B,C,D$in$20=(A+C)s^3+(B+D)s^2+16As+16B$

$\frac{20}{s^2\left(s^2+16\right)}=\frac{5}{4s^2}-\frac{5}{4\left(s^2+16\right)}$

Hence the coefficient is$\frac{5}{4s^2}-\frac{5}{4\left(s^2+16\right)}$

Decomposing using partial fraction technique.

Partial fraction technique is used for composition of the given equation.

$$\begin{gathered} \star \frac{100s-20}{s^2\left(s^2+16\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+16} \\ =\frac{As\left(s^2+16\right)+B\left(s^2+16\right)+(Cs+D)s^2}{s^2\left(s^2+16\right)} \end{gathered}$$

Compare the coefficients.

$$\begin{gathered} A+C=0 \\ B+D=0 \\ 16A=100 \\ 16B=20 \\ C=A \\ D=B \\ A=\left(100\right)/\left(16\right) \\ B=\left(5\right)/\left(4\right) \\ C=\left(25\right)/\left(4\right) \\ D=\left(5\right)/\left(4\right) \\ A=\left(25\right)/\left(4\right) \\ B=\left(5\right)/\left(4\right) \end{gathered}$$

Substitute the value of in $\frac{As\left(s^2+16\right)+B\left(s^2+16\right)+(Cs+D)s^2}{s^2\left(s^2+16\right)}$

$\frac{100s-20}{s^2\left(s^2+16\right)}=\frac{25}{4s}-\frac{5}{4s^2}+\frac{5-25s}{4\left(s^2+16\right)}$

The decomposed value of $\frac{100s-20}{s^2\left(s^2+16\right)}is\frac{25}{4s}-\frac{5}{4s^2}+\frac{5-25s}{4\left(s^2+16\right)}$

Take inverse Laplace on both sides of the equation.

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{X\right(s\left)\right\}=\frac{5}{4}{\mathcal{L}}^{-1}\left\{\frac{1}{s^2}\right\}-\frac{5}{4}\frac{1}{4}{\mathcal{L}}^{-1}\left\{\frac{4}{s^2+4^2}\right\}+\frac{25}{4}{\mathcal{L}}^{-1}\left\{e^{-5s}\frac{1}{s}\right\} \\ -\frac{5}{4}{\mathcal{L}}^{-1}\left\{e^{-5s}\frac{1}{s^2}\right\}-\frac{25}{4}{\mathcal{L}}^{-1}\left\{\frac{e^{-5s}s}{s^2+4^2}\right\}+\frac{5}{4}\frac{1}{4}{\mathcal{L}}^{-1}\left\{\frac{e^{-5s}4}{s^2+4^2}\right\} \end{gathered}$$

role="math" localid="1663943366725" $$\begin{gathered} \frac{5}{4}t-\frac{5}{16}\sin \left(4t\right)+\frac{25}{4}U(t-5)-\frac{5}{4}(t-5)U(t-5) \\ -\frac{25}{4}\cos \left(4\right(t-5\left)\right)U(t-5)+\frac{5}{16}\sin \left(4\right(t-5\left)\right)U(t-5) \\ {\mathcal{L}}^{-1}\left\{\frac{n!}{s^{n+1}}\right\}=t^n,{\mathcal{L}}^{-1}\left\{\frac{k}{s^2+k^2}\right\}=\sin kt \end{gathered}$$

So,

$$\begin{gathered} x\left(t\right)=\frac{5}{4}t-\frac{5}{16}\sin \left(4t\right)+\frac{25}{4}U(t-5)-\frac{5}{4}(t-5)U(t-5) \\ -\frac{25}{4}\cos \left(4\right(t-5\left)\right)U(t-5)+\frac{5}{16}\sin \left(4\right(t-5\left)\right)U(t-5) \end{gathered}$$

Hence the inverse. The Laplace transform is

$$\begin{gathered} x\left(t\right)=\frac{5}{4}t-\frac{5}{16}\sin \left(4t\right)+\frac{25}{4}U(t-5)-\frac{5}{4}(t-5)U(t-5) \\ -\frac{25}{4}\cos \left(4\right(t-5\left)\right)U(t-5)+\frac{5}{16}\sin \left(4\right(t-5\left)\right)U(t-5) \end{gathered}$$

Graphing the solution

The sketch of the solution curve drawn using graphing calculator is shown below.